Circumcircle of a Triangle

A triangle is a polygon with 3 sides. A triangle is more special as compared to other polygons as it is the polygon having the least number of sides. A triangle has six main elements, three sides, and three angles. A circumcircle is a circle formed using the vertices of the triangle. In real life, we use circumcircle in coastal navigation. It is also a way of obtaining a position line using a sextant when no compass is available.

In this article, we will cover the concept of the Circumcircle. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of six questions have been asked on this concept including one in 2021.

What is the Circumcircle of a Triangle?

The circumcircle of triangle $A B C$ is the unique circle passing through the three vertices $A, B,$ and $C$. Its center, the circumcenter $O$, is the intersection of the perpendicular bisectors of the three sides. The circumradius is always denoted by $R$.

Radius of Circumcircle

The radius of the circumcircle of a $ΔABC, R$ is given by the law of sines:

$\mathrm{R}=\frac{a}{2 \sin \mathrm{A}}=\frac{b}{2 \sin \mathrm{B}}=\frac{c}{2 \sin \mathrm{C}}$

Where $a, b, c$ is the side length of a triangle.

Derivation of Radius of Circumcircle

The perpendicular bisector of the sides $A B, B C$, and $C A$ intersects at point $O$. So, $O$ is the circumcentre and

$
O A=O B=O C=R
$
Let $D$ be the midpoint of $B C$.

$
\begin{aligned}
& \angle B O C=2 \angle B A C=2 A \\
& \angle B O D=\angle C O D=A
\end{aligned}
$
So, in $\triangle \mathrm{OBD}$,

$
\begin{aligned}
& \sin \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{OB}}=\frac{a / 2}{\mathrm{R}}=\frac{a}{2 \mathrm{R}} \\
& \Rightarrow \mathrm{R}=\frac{a}{2 \sin \mathrm{A}}
\end{aligned}
$

similarly,

$
\mathrm{R}=\frac{b}{2 \sin \mathrm{B}} \text { and } \mathrm{R}=\frac{c}{2 \sin \mathrm{C}}
$
R can also be written in terms of the area of the triangle

Relation between circumradius and Area of triangle

Area of $ΔABC,$

$
\begin{aligned}
& \Delta=\frac{1}{2} b \cdot \mathrm{c} \sin \mathrm{A} \\
& \Rightarrow \sin \mathrm{A}=\frac{2 \Delta}{b c}
\end{aligned}
$

and, $\mathrm{R}=\frac{a}{2 \sin \mathrm{A}}$
From (i) and (ii)

$
\mathrm{R}=\frac{a b c}{4 \Delta}
$

Note:

1) The circumcentre of the Acute angled triangle lies inside the triangle.

2) The circumcentre of the obtuse-angled triangle lies outside the triangle.

3) The circumcentre of the right angle triangle is the mid-point of its hypotenuse.

Solved Examples Based on Circumcircle

Example 1: Let the centroid of an equilateral triangle $ABC$ be at the origin. Let one of the sides of the equilateral triangle be along the straight line $x + y =3$. If $R$ and $r$ be the radius of the circumcircle and incircle respectively of $\triangle A B C$ then $(R+r)$ is equal to [JEE MAINS 2021]

Solution

$\begin{aligned}
& \mathrm{r}=\mathrm{OM}=\frac{3}{\sqrt{2}} \\
& \sin 30^{\circ}=\frac{1}{2}=\frac{\mathrm{r}}{\mathrm{R}} \\
& \Rightarrow \mathrm{R}=\frac{6}{\sqrt{2}} \\
& \therefore \mathrm{r}+\mathrm{R}=\frac{9}{\sqrt{2}}
\end{aligned}
$

Hence, the answer is $\frac{9}{\sqrt{2}}$

Example 2: If the area of a triangle is $1$ and the products of sides $abc = 8$, then the radius of its circumcircle is.

Solution: We know that,

$\begin{aligned} & R=\frac{a b c}{4 \Delta} \\ & \mathrm{R}=\frac{8}{4.1}=2\end{aligned}$

Hence, the answer is $2$.

Example 3: In triangle ABC, a,b, and c are in G.P., b=2, and the radius of the circumcircle of this triangle is R=1 then find the area of triangle ABC.

Solution: We know that

$
\begin{aligned}
& \Delta=\frac{a b c}{4 R} \\
& \text { Also } a, b, c \text { in } G \cdot P . \Rightarrow b^2=a c
\end{aligned}
$

and $R=1$
So,

$
\Delta=\frac{a c b}{4.1}=\frac{b^3}{4}=\frac{8}{4}=2
$

Hence, the answer is $2$

Example 4: $\ln \Delta \mathrm{ABC}$, let $A D$ be the median and $O, G$, and $P$ be respectively the circumcentre, centroid, and orthocentre. Then $\triangle O G D$ is directly similar to

Solution

Trigonometric Ratios of Functions

$\begin{aligned} & \cos \theta=\frac{\text { Base }}{H y p} \\ & \tan \theta=\frac{O p p}{\text { Base }}\end{aligned}$

From the figure, we can observe that $\Delta O G D$ is directly similar to $\Delta P G A$
Hence, the answer is $\triangle \mathrm{PGA}$

Example 5: Let $(5, \mathrm{a} / 4)$ be the circumcenter of a triangle with vertices $\mathrm{A}(\mathrm{a},-2), \mathrm{B}(\mathrm{a}, 6)$ and $\mathrm{C}(\mathrm{a} / 4,-2)$. Let $\alpha$ denote the circumradius, $\beta$ denote the area, and $\gamma$ denote the perimeter of the triangle. Then $\alpha+\beta+\gamma$ is

Solution: $D$ is the midpoint of $A B . A B$ is perpendicular to $P D$

$
2=\mathrm{a} / 4 \Rightarrow \mathrm{a}=8
$

$\begin{aligned} & \mathrm{A}(8,-2), \mathrm{B}(8,6), \mathrm{C}(2,-2), \mathrm{P}(5,2) \\ & \alpha=\mathrm{AP}=\sqrt{9+16}=5 \\ & \mathrm{AP}=8, \mathrm{BC}=10, \mathrm{AC}=6 \\ & \text { Area }=\beta=(1 / 2) \times 8 \times 6=24 \\ & \text { Perimeter }=\gamma=24 \\ & \alpha+\beta+\gamma=5+24+24=53\end{aligned}$

Hence, the answer is $53.$

Summary

The circle passing through the vertices of a triangle is called a circumcircle. The circumcenter $O$, is the intersection of the perpendicular bisectors of the three sides. Apart from mathematics, this concept of circumcenter is used in many real life problems in Engineering etc.

Frequently Asked Questions (FAQs)

Q1) what is circumcircle?

Answer: The circle passing through the vertices of a triangle is called a circumcircle. The circumcircle of triangle $ABC$ is the unique circle passing through the three vertices $A, B,$ and $C$.

Q2) What is circumcentre?

Answer: The intersection of the perpendicular bisectors of the three sides is called circumcentre. It is denoted by the letter '$O$'.

Q3) Where does the circumcentre of the obtuse angle triangle lie?

Answer: The circle passing through the vertices of a triangle is called a circumcircle. The circumcentre of the obtuse-angled triangle lies outside the triangle.

Q4) What is the distance of the circumcentre from all the vertices of the triangle?

Answer: The intersection of the perpendicular bisectors of the three sides is called circumcentre The distance of the circumcentre from all the vertices of the triangle is the same which is equal to the circumradius of the triangle.

Q5 ) How to calculate the radius of a circumcircle?

Answer: The radius of the circumcircle of a $ΔABC, R$ is given by the law of sines:

$\mathbb{R}=\frac{a}{2 \sin \mathrm{A}}=\frac{b}{2 \sin \mathrm{B}}=\frac{c}{2 \sin \mathrm{C}}$